📐 Eigenvalues as Roots

$\\lambda$ is an eigenvalue of $A$ if and only if $\\lambda$ is a root of the characteristic polynomial.

Proof: $\\lambda$ is an eigenvalue $\\iff$ $(A - \\lambda I)v = 0$ has a nonzero solution

$\\iff$ $A - \\lambda I$ is not invertible

$\\iff$ $\\det(A - \\lambda I) = 0$

$\\iff$ $\\det(\\lambda I - A) = 0$

$\\iff$ $\\lambda$ is a root of the characteristic polynomial.

From: Advanced Linear Algebra

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